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Solving Trig Problems with Multiple Angles – General Solutions
Now let’s solve the same multiple angle problems, but get solutions between 0 and \(2\pi k\).
Note again that when we solve these types of trig problems, we always want to solve for the General Solution first, and then go back and see how many solutions are on
the Unit Circle (between 0 and \(2\pi k\)).
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Solving Trig Problems with Multiple Angles: \(\boldsymbol{\left[ {0,2\pi } \right]}\)
\(\displaystyle \tan \left( {3\theta } \right)=\sqrt{3}\)
\(3\theta =\frac{\pi }{3}+\pi k\,\,\,\,\,\,\,\,3\theta
=\frac{{4\pi }}{3}+\pi k\)
Looking at Unit Circle, this can be simplified to:
\(\displaystyle \begin{array}{c}3\theta =\frac{\pi }
{3}+\pi k\\\,\theta =\frac{\pi }{9}+\frac{\pi }
{3}k\,\,=\,\,\frac{\pi }{9}+\frac{{3\pi }}{9}k\end{array}\)
Looking at the Unit Circle, we can get solutions
between 0 and \(2\pi \) by adding \(\frac{{3\pi }}{9}\),
staying under \(2\pi \):
\(\theta =\left\{ {\frac{\pi }{9},\frac{{4\pi }}
{9},\frac{{7\pi }}{9},\frac{{10\pi }}{9},\frac{{13\pi }}
{9},\frac{{16\pi }}{9}} \right\}\)
\(\displaystyle 2\cos \left( {3x} \right)+\sqrt{3}=0\)
\(\displaystyle \begin{array}{c}\cos \left( {3x} \right)=-
\frac{{\sqrt{3}}}{2}\\\,\,\,3x=\frac{{5\pi }}{6}+2\pi
k\,\,\,\,\,\,\,\,\,3x=\frac{{7\pi }}{6}+2\pi k\\\,\,\,x=\frac{{5\pi }}
{{18}}+\frac{{12}}{{18}}\pi k\,\,\,\,\,\,\,\,\,\,x=\frac{{7\pi }}
{{18}}+\frac{{12}}{{18}}\pi k\end{array}\)
Looking at the Unit Circle, we can get solutions between 0 and
\(2\pi \) by adding \(\frac{{12\pi }}{{18}}\), staying under \(2\pi \):
\(x=\left\{ {\frac{{5\pi }}{{18}},\frac{{17\pi }}{{18}},\frac{{29\pi }}
{{18}},\frac{{7\pi }}{{18}},\frac{{19\pi }}{{18}},\frac{{31\pi }}{{18}}}
\right\}\)
\(\displaystyle \sqrt{2}\sec \left( {\frac{x}{6}} \right)-
2=0\)
\(\displaystyle \begin{array}{c}\sec \left( {\frac{x}{6}}
\right)=\frac{2}{{\sqrt{2}}}\,\,\,\,\,\,\,\left( {\cos \left(
{\frac{x}{6}} \right)=\frac{{\sqrt{2}}}{2}}
\right)\\\,\,\frac{x}{6}=\frac{\pi }{4}+2\pi
k\,\,\,\,\,\,\,\,\,\,\,\frac{x}{6}=\frac{{7\pi }}{4}+2\pi
k\\\,\,\,x=\frac{{6\pi }}{4}+12\pi
k\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{42\pi }}{4}+12\pi
k\end{array}\)
Looking at the Unit Circle, the only solution that will
work is \(\frac{{6\pi }}{4}=\frac{{3\pi }}{2}\).
\(x=\left\{ {\frac{{3\pi }}{2}} \right\}\)
\(\displaystyle 5{{\cos }^{2}}\left( {\frac{\theta }{3}}
\right)=5\)
\(\displaystyle \begin{array}{c}\sqrt{{{{{\cos
}}^{2}}\left( {\frac{\theta }{3}}
\right)}}=\sqrt{1}\\cos\left( {\frac{\theta }{3}}
\right)=\pm 1\\\,\,\,\frac{\theta }{3}=0+2\pi
k\,\,\,\,\,\,\,\,\,\,\frac{\theta }{3}=\pi +2\pi
k\end{array}\)
\(\displaystyle 2{{\sin }^{2}}\left( {2x} \right)=1\)
\(\displaystyle \begin{array}{c}\sqrt{{{{{\sin }}^{2}}\left( {2x}
\right)}}=\sqrt{{\frac{1}{2}}}\\\sin \left( {2x} \right)=\,\,\pm \frac{1}
{{\sqrt{2}}}=\,\,\pm \frac{{\sqrt{2}}}{2}\\\,\,\,2x=\frac{\pi }{4}+2\pi
k\,\,\,\,\,\,\,\,\,\,2x=\frac{{3\pi }}{4}+2\pi k\\\,\,\,2x=\frac{{5\pi }}
{4}+2\pi k\,\,\,\,\,\,\,\,2x=\frac{{7\pi }}{4}+2\pi k\end{array}\)
Looking at the Unit Circle, this can be simplified to:
\(\displaystyle \begin{array}{c}2x=\frac{\pi }{4}+\pi
k\,\,\,\,\,\,\,\,\,\,2x=\frac{{3\pi }}{4}+\pi k\\\,\,\,\,\,\,\,\,x=\frac{\pi }
\(\displaystyle \tan \left( {\frac{\theta }{2}-\frac{\pi }{3}}
\right)=-1\)
\(\displaystyle \begin{array}{c}\frac{\theta }{2}-
\frac{\pi }{3}=\frac{{3\pi }}{4}+\pi k\\\frac{\theta }{2}-
\frac{\pi }{3}=\frac{{7\pi }}{4}+\pi k\end{array}\)
Looking at the Unit Circle, this can be simplified to:
\(\displaystyle \begin{array}{c}\frac{\theta }
{2}-\frac{\pi }{3}=\frac{{3\pi }}{4}+\pi
k\\\frac{\theta }{2}=\left( {\frac{{3\pi }}
{4}+\frac{\pi }{3}} \right)+\,\,\pi
k\\\frac{\theta }{2}=\frac{{13\pi }}{{12}}+\pi
Looking at the Unit Circle, this can be simplified to:
\(\displaystyle \begin{array}{c}\frac{\theta }{3}=\pi
k\\\theta =3\pi k\end{array}\)
Looking at the Unit Circle, the only solution that will
work is 0.
\(\displaystyle \theta =\left\{ 0 \right\}\)
{8}+\frac{\pi }{2}k,\,\,\,x=\frac{{3\pi }}{8}+\frac{\pi }
{2}k\end{array}\)
Looking at the Unit Circle, we can get solutions between 0 and
\(2\pi \) by adding \(\frac{{4\pi }}{8}\), staying under \(2\pi \):
\(\displaystyle x=\left\{ \begin{array}{l}\frac{\pi }{8},\frac{{5\pi }}
{8},\frac{{9\pi }}{8},\frac{{13\pi }}{8},\\\frac{{3\pi }}{8},\frac{{7\pi }}
{8},\frac{{11\pi }}{8},\frac{{15\pi }}{8}\end{array} \right\}\)
k\\\theta =\frac{{13\pi }}{6}+2\pi
k\end{array}\)
Looking at the Unit Circle, the only solution that will
work is:
\(\begin{array}{c}\frac{{13\pi }}{6}-2\pi
=\frac{\pi }{6}\\\theta =\left\{ {\frac{\pi }{6}}
\right\}\end{array}\)
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